Question

Subtract.
(i) $3\frac{1}{3}-1\frac{1}{4}$
(ii) $5\frac{1}{2}-3\frac{1}{3}$
(iii) $7\frac{1}{8}-6\frac{1}{10}$
(iv) $7\frac{1}{2}-3\frac{1}{5}$

Answer 1

(i)

$3\frac{1}{3}-1\frac{1}{4}$

$=\frac{10}{3}-\frac{5}{4}$

LCM of $3,4$ is $12$.

$=\frac{10}{3}\times \frac{4}{4}-\frac{5}{4}\times \frac{3}{3}$

$=\frac{40}{12}-\frac{15}{12}$

$=\frac{25}{12}$

$=2\frac{1}{12}$

$\therefore 3\frac{1}{3}-1\frac{1}{4}=2\frac{1}{12}$

(ii)

$5\frac{1}{2}-3\frac{1}{3}$

$=\frac{11}{2}-\frac{10}{3}$

LCM of $2,3$ is $6$.

$=\frac{11}{2}\times \frac{3}{3}-\frac{10}{3}\times \frac{2}{2}$

$=\frac{33}{6}-\frac{20}{6}$

$=\frac{33-20}{6}$

$=\frac{13}{6}$

$=2\frac{1}{6}$

$\therefore 5\frac{1}{2}-3\frac{1}{3}=2\frac{1}{6}$
(iii)

$7\frac{1}{8}-6\frac{1}{10}$

$=\frac{57}{8}-\frac{61}{10}$

LCM of $8,10$ is $40$.

$=\frac{57}{8}\times \frac{5}{5}-\frac{61}{10}\times \frac{4}{4}$

$=\frac{285}{40}-\frac{244}{40}$

$=\frac{41}{40}$

$=1\frac{1}{40}$

$\therefore 7\frac{1}{8}-6\frac{1}{10}=1\frac{1}{40}$
(iv)

$7\frac{1}{2}-3\frac{1}{5}$

$=\frac{15}{2}-\frac{16}{5}$

LCM of $2,5$ is $10$.

$=\frac{15}{2}\times \frac{5}{5}-\frac{16}{5}\times \frac{2}{2}$

$=\frac{75}{10}-\frac{32}{10}$

$=\frac{75-32}{10}$

$=\frac{43}{10}$

$=4\frac{3}{10}$

$\therefore 7\frac{1}{2}-3\frac{1}{5}=4\frac{3}{10}$